Question 161: Thermodynamics & KTG

Question

The temperature of a gas is $-78^{\circ} \mathrm{C}$ and the average translational kinetic energy of its molecules is $\mathrm{K}$. The temperature at which the average translational kinetic energy of the molecules of the same gas becomes $2 \mathrm{~K}$ is :

Accepted Answer

The average translational kinetic energy ($K_{avg}$) of a molecule is directly proportional to the absolute temperature (T) of the gas, as described by the equation:
$K_{avg} = \frac{3}{2}kT$
Where:
$K_{avg}$ is the average kinetic energy,
$k$ is the Boltzmann's constant, and
$T$ is the temperature in Kelvin.
From the given problem, if the temperature of the gas is $-78^{\circ}C$, which in Kelvin is $T_1 = -78 + 273 = 195K$, and the average translational kinetic energy is $K$, when the energy becomes $2K$, we need to find the new temperature $(T_2)$.
Using the direct proportionality relation, we can set up the following equation:
$K \propto T$
$2K = K_2 = \frac{3}{2}kT_2$
Given that at $T_1$, the kinetic energy is $K$, and at $T_2$, it's $2K$, we can use the ratio as follows:
$\frac{K_2}{K_1} = \frac{2K}{K} = 2 = \frac{T_2}{T_1}$
Thus, we have:
$T_2 = 2T_1 = 2 	imes 195K = 390K$
To find the temperature in Celsius, we convert $390K$ back to Celsius:
$T_{2(Celsius)} = 390 - 273 = 117^{\circ}C$
Therefore, the temperature at which the average translational kinetic energy of the molecules of the same gas becomes $2K$ is $117^{\circ}C$. The correct answer is:
Option D $117^{\circ} \mathrm{C}$

Answer

$-78^{\circ} \mathrm{C}$

Answer

$127^{\circ} \mathrm{C}$

Answer

$-39^{\circ} \mathrm{C}$

Answer

$117^{\circ} \mathrm{C}$

Thermodynamics & KTG

The temperature of a gas is -78^{\circ} \mathrm{C} and the average translational kinetic energy of its molecules is \mathrm{K}. The temperature at which the average translational kinetic energy of the molecules of the same gas becomes 2 \mathrm{~K} is :

Choose the correct answer: